Answer:
6.13 s
219 N
Explanation:
Newton's law in the x direction:
∑F = ma
150 cos 30° N − 50 N = (30 kg) a
a = 2.66 m/s²
Δx = v₀ t + ½ at²
(50 m) = (0 m/s) t + ½ (2.66 m/s²) t²
t = 6.13 s
Newton's law in the y direction:
∑F = ma
Fn + 150 sin 30° N − (30 kg) (9.8 m/s²) = 0
Fn = 219 N
Answer:
The additional trials needed is 48 trials
Explanation:
Given;
initial number of trials, n = 16 trials
the standard deviation, σ = 0.24 s
initial standard error, ε = 0.06 s
The standard error is given by;
To reduce the standard error to 0.03 s, let the additional number of trials = x
Therefore, the additional trials needed is 48 trials.
The answer to your question is aluminum.
Answer:
≈19.144°C.
Explanation:
all the details are in the attachment.
Note, that c₁, m₁, t₁ are the parameters of the sample of brass; c₂, m₂ and t₂ are the parameters of the sample of water.
P.S. change the provided design according Your requirements.
Answer:
Time is 14.8 s and cannot landing
Explanation:
This is a kinematic exercise with constant acceleration, we assume that the acceleration of the jet to take off and landing are the same
Calculate the time to stop, where it has zero speed
Vf² = Vo² + a t
t = - Vo² / a
t = - 110²/(-7.42)
t = 14.8 s
This is the time it takes to stop the jet
Let's analyze the case of the landing at the small airport, let's look for the distance traveled to land, where the speed is zero
Vf² = Vo² + 2 to X
X = -Vo² / 2 a
X = -110² / 2 (-7.42)
X = 815.4 m
Since this distance is greater than the length of the runway, the jet cannot stop
Let's calculate the speed you should have to stop on a track of this size
Vo² = 2 a X
Vo = √ (2 7.42 800)
Vo = 109 m / s
It is conclusion the jet must lose some speed to land on this track
